Question

A man weighing 160 lb stands on the floor of an elevator. What force will be exerted on floor if it (i) descends with a uniform acceleration of 1 ft/sec 2 (pi) ascends with a uniform acceleration of 1 ft/sec 2

Answer 1

(a) When elevator descends weight will be 155.026 lb

(b) When elevator ascends weight will be 164.954 lb

Step-by-step explanation:

It is given weight of man on the floor of elevator F = 160 lb

Acceleration due to gravity
`g=32.17ft/sec^2`

At the floor of elevator weight is equal to
`W=mg`

So
`160=m* 32.17`

m =
`4.973lbsec^2/ft`

(ii) When the elevator descends with uniform acceleration
`1ft/sec^2`

Weight will be equal to
`W=m(g+a)=4.973* (32.17+1)=155.026lb`

(ii) When elevator is moving upward

Weight is equal to
`W=m(g+a)=4.973* (32.17+1)=164.954lb`